CS 441 Lecture, Dr. Lawlor

Ordinary integers can only represent integral values.  "Floating-point numbers" can represent non-integral values.   This is useful for engineering, science, statistics, graphics, and any time you need to represent numbers from the real world, which are rarely integral!

Floats store numbers in an odd way--they're really storing the number in scientific notation, like
    x = + 3.785746 * 105
Note that:
Scientific notation is designed to be compatible with slide rules (here's a circular slide rule demo); slide rules are basically a log table starting at 1.  This works because log(1) = 0, and log(a) + log(b) = log(ab).  But slide rules only give you the mantissa; you need to figure out the exponent yourself.  The "order of magnitude" guess that engineers (and I) like so much is just a calculation using zero significant digits--no mantissa, all exponent.

Normalized Numbers

Scientific notation can represent the same number in several different ways:
    x = + 3.785746 * 105  = + 0.3785746 * 106 = + 0.003785746 * 107 = + 37.85746 * 104 

It's common to "normalize" a number in scientific notation so that:
  1. There's exactly one digit to the left of the decimal point.
  2. And that digit ain't zero.
This means the 105 version is the "normal" way to write the number above.

In binary, a "normalized" number *always* has a 1 at the left of the decimal point (if it ain't zero, it's gotta be one).  So there's no reason to even store the 1; you just know it's there!

(Note that there are also "denormalized" numbers, like 0.0, that don't have a leading 1.  This is how zero is represented--there's an implicit leading 1 only if the exponent field is nonzero, an implicit leading 0 if the exponent field is zero...)

Float as Bits

Floats represent continuous values.  But they do it using discrete bits.

A "float" (as defined by IEEE Standard 754) consists of three bitfields:
Fraction (or "Mantissa")
1 bit--
  0 for positive
  1 for negative
8 unsigned bits--
  127 means 20
  137 means 210
23 bits-- a binary fraction.

Don't forget the implicit leading 1!
The sign is in the highest-order bit, the exponent in the next 8 bits, and the fraction in the remaining bits.

The hardware interprets a float as having the value:

    value = (-1) sign * 2 (exponent-127) * 1.fraction

Note that the mantissa has an implicit leading binary 1 applied (unless the exponent field is zero, when it's an implicit leading 0; a "denormalized" number).

For example, the value "8" would be stored with sign bit 0, exponent 130 (==3+127), and mantissa 000... (without the leading 1), since:

    8 = (-1) 0 * 2 (130-127) * 1.0000....

You can actually dissect the parts of a float using a "union" and a bitfield like so:
/* IEEE floating-point number's bits:  sign  exponent   mantissa */
struct float_bits {
unsigned int fraction:23; /**< Value is binary 1.fraction ("mantissa") */
unsigned int exp:8; /**< Value is 2^(exp-127) */
unsigned int sign:1; /**< 0 for positive, 1 for negative */

/* A union is a struct where all the fields *overlap* each other */
union float_dissector {
float f;
float_bits b;

float_dissector s;
std::cout<<s.f<<"= sign "<<s.b.sign<<" exp "<<s.b.exp<<" fract "<<s.b.fraction<<"\n";
return 0;
(Executable NetRun link)

There are several different sizes of floating-point types:
C Datatype
Approx. Precision
Approx. Range
Exponent Bits
Fraction Bits
+-1 range
4 bytes (everywhere)
8 bytes (everywhere)
long double
12-16 bytes (if it exists)

Nowadays floats have roughly the same performance as integers: addition takes a little over a nanosecond (slightly slower than integer addition); multiplication takes a few nanoseconds; and division takes a dozen or more nanoseconds.  That is, floats are now cheap, and you can consider using floats for all sorts of stuff--even when you don't care about fractions.

Roundoff in Representation

0.1 decimal is an infinitely repeating pattern in binary (0.0(0011), with 0011 repeating).  This means multiplying by some *finite* pattern to approximate 0.1 is only an approximation of really dividing by the integer 10.0.  The exact difference is proportional to the precision of the numbers and the size of the input data:
for (int i=1;i<1000000000;i*=10) {
double mul01=i*0.1;
double div10=i/10.0;
double diff=mul01-div10;
std::cout<<"i="<<i<<" diff="<<diff<<"\n";
(executable NetRun link)

On my P4, this gives:
i=1  diff=5.54976e-18
i=10 diff=5.55112e-17
i=100 diff=5.55112e-16
i=1000 diff=5.55112e-15
i=10000 diff=5.55112e-14
i=100000 diff=5.55112e-13
i=1000000 diff=5.54934e-12
i=10000000 diff=5.5536e-11
i=100000000 diff=5.54792e-10
Program complete. Return 0 (0x0)
That is, there's a factor of 10^-15 difference between double-precision 0.1 and the true 1/10!

Roundoff in Arithmetic

They're funny old things, floats.  The fraction part only stores so much precision; further bits are lost.  For example, in reality,
    1.2347654 * 104 = 1.234* 104 + 7.654* 100
But to three decimal places,
    1.234 * 104 = 1.234* 104 + 7.654* 100
which is to say, adding a tiny value to a great big value might not change the great big value at all, because the tiny value gets lost when rounding off to 3 places.  This "roundoff" has implications.

For example, adding one repeatedly will eventually stop doing anything:
float f=0.73;
while (1) {
volatile float g=f+1;
if (g==f) {
printf("f+1 == f at f=%.3f, or 2^%.3f\n",
return 0;
else f=g;
(executable NetRun link)
Recall that for integers, adding one repeatedly will *never* give you the same value--eventually the integer will wrap around, but it won't just stop moving like floats!

For another example, floating-point arithmetic isn't "associative"--if you change the order of operations, you change the result (up to roundoff):
    1.2355308 * 104 = 1.234* 104 + (7.654* 100 + 7.654* 100)
    1.2355308 * 104 = (1.234* 104 + 7.654* 100) + 7.654* 100
In other words, parenthesis don't matter if you're computing the exact result.  But to three decimal places,
    1.235 * 104 = 1.234* 104 + (7.654* 100 + 7.654* 100)
    1.234 * 104 = (1.234* 104 + 7.654* 100) + 7.654* 100
In the first line, the small values get added together, and together they're enough to move the big value.  But separately, they splat like bugs against the windshield of the big value, and don't affect it at all!
double lil=1.0;
double big=pow(2.0,64);
printf(" big+(lil+lil) -big = %.0f\n", big+(lil+lil) -big);
printf("(big+lil)+lil -big = %.0f\n",(big+lil)+lil -big);
(executable NetRun link)
float gnats=1.0;
volatile float windshield=1<<24;
float orig=windshield;
for (int i=0;i<1000;i++)
windshield += gnats;

if (windshield==orig) std::cout<<"You puny bugs can't harm me!\n";
else std::cout<<"Gnats added "<<windshield-orig<<" to the windshield\n";
(executable NetRun link)

In fact, if you've got a bunch of small values to add to a big value, it's more roundoff-friendly to add all the small values together first, then add them all to the big value:
float gnats=1.0;
volatile float windshield=1<<24;
float orig=windshield;
volatile float gnatcup=0.0;
for (int i=0;i<1000;i++)
gnatcup += gnats;
windshield+=gnatcup; /* add all gnats to the windshield at once */

if (windshield==orig) std::cout<<"You puny bugs can't harm me!\n";
else std::cout<<"Gnats added "<<windshield-orig<<" to the windshield\n";
(executable NetRun link)

Roundoff is very annoying, but it doesn't matter if you don't care about exact answers, like in simulation (where "exact" means the same as the real world, which you'll never get anyway) or games.

Denormal Numbers

A denormal (tiny) number doesn't have an implicit leading 1; it's got an implicit leading 0.  This means the ciruitry for doing arithmetic on denormal numbers doesn't look much like the circuitry for doing arithmetic on ordinary normalized numbers.  On modern machines, there *is* no circuitry for computing on denormal numbers--instead, the hardware traps out to software when it hits a denormal value.

This would be fine, except software is way slower than hardware--about 25x slower on the NetRun machine!
float f=pow(2,-128); // denormal

int foo(void) {
f*=1.00001; //<- you can do almost any operation here...
return 0;
(executable NetRun link)

As written, this takes 328ns per execution of foo, which is crazy slow.

If you initialize f to, say, 3.0 (or any non-denormal value), this takes like 13ns per execution, which is reasonable.

That is, floating-point code can take absurdly longer when computing denormals (infinities have the same problem).  Denormals can have a huge impact on the performance of real code--I've written a paper on this.

The easiest way to fix denormals is to round them off to zero--one trick for doing this is to just add a big value ("big" compared to a denormal can be like 1.0e-10) and then subtract it off again!